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MATH 150-768 Elementary Statistics – Final Review (Spring 2012)
Chapter 2. Summarizing and Graphing Data
1. For the data given below,
70
61
74
80
58
66
53
70
65
71
63
60
75
62
68
a. Complete constructing a Frequency distribution by filling in the frequencies
Grades
51 – 55
56 – 60
61 – 65
66 – 70
71 – 75
76 – 80
Frequency
Ans. 1
Ans. 2
Ans. 4
Ans. 4
Ans. 3
Ans. 1
b. Plot the histogram of the data, in the space below, using the above frequency
distribution. Ans. See graph
Number of TV Sets
53
58
63 68 73
78
c. Does the frequency distribution in the above problem appear to have a normal
distribution? Why? Ans. Yes. Appear to have a normal distribution.
d. Identify the class width, class midpoint, class boundaries, lower and upper class limits for
the given frequency distribution.
Class width: Ans. 5
Class midpoints: Ans. 53, 58, 63, 68, 73, 78
Class boundaries:
Ans. 50.5, 55.5, 60.5, 65.5, 70.5, 75.5, 80.5
e. Construct the Stem-and-leaf plot of the original data.
Ans.
Stem Leaves
5
38
6
0123568
7
00145
8
0
1
Chapter 3.
2. A student obtains the following grades in 8 quizzes.
65
75
80
85
70
90
Find the following:
1) Mean
Ans. 76.3
2) Median
Ans. 75
3) Mode Ans. 70, 75 (bi-modal)
4) Standard Deviation Ans. 8.3
5) Variance
Ans. 69.6
70
75
3. Determine which score corresponds to the higher relative position: a score of 51.5 on a test
for which = 47 and s = 9, a score of 5.9 on a test for which = 4.2 and s = 1.2 or a score of
460.8 on a test for which = 444 and s = 42.
Ans. a score of 5.9 on a test for which = 4.2 and s = 1.2 (z-scores are, 0.5, 1.4, 0.4)
4. Find the percentile for the data value 6:
4, 12, 9, 6, 4, 4, 12, 6, 4, 12, 2, 12, 15, 5, 9, 4, 12, 9, 6, 12
Ans. the percentile of the data value 6 is 35. (i.e. 35% of the data is less than 6)
5. The following is a table for the grades in an Elementary Algebra test. Construct the boxplot.
70
82
74
80
58
66
63
70
85
75
80
70
75
62
80
Ans. (min = 58, Q1 = 62, Q2 = 74, Q3 = 80, max = 85)
74
58
62
55
60
85
80
65
70 75
80 85
90
Chapter 4. Probability
6. A batch consists of 12 defective coils and 88 ones. Find the probability of getting two good
coils when two coils are randomly selected if the first selection is replaced before the second
is made. Ans. (0.88)(0.88) = 0.7744
(independent)
7. A sample of 3 different calculators is randomly selected from a group containing 41 that are
defective and 22 that have no defects. What is the probability that all four of the calculators
selected are defective? Round to four decimal places.
Ans. (414039)/(636261) = 0.2684
(dependent)
Provide the written description of the complement of the given event.
8. Of ten adults, at least one of them has high blood pressure.
Ans. out of ten adults, none of them had high blood pressure.
Chapter 5 Discrete Probability Distributions
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9. Assume that a procedure yields a binomial distribution with a trial repeated n times. Use the
binomial probability formula to find the probability of x successes given the probability p of
success on a single trial. Round to the three decimal places.
n = 4, x = 3, p = 1/6
Ans. 0.015
Find the indicated probability
10. The brand name of a certain chain of coffee shops has 53% recognition rate in the town of
Coffleton. An executive from the company wants to verify the recognition rate as the
company is interested in opening a coffee shop in the town. He selects a random sample of 7
Coffleton residents. Find the probability that exactly 4 of 7 Coffleton residents recognize the
brand name.
Ans. 0.287
Use the given values of n and p to find the minimum usual value  – 2 and the maximum usual
value  + 2. Round your answer to the nearest hundredth unless otherwise noted.
11. n = 104; p = 0.21 Ans. min usual = 13.53, max usual = 30.15
Chapter 6 (6.2 – 6.5)
12. Find the area of the shaded region. The graph depicts IQ scores of adults, and those scores
are normally distributed with a mean of 100 and a standard deviation of 15.
Ans. 0.7938
85
100
125
Solve the problem.
13. A final exam in Math160 has a mean of 73 and a standard deviation of 7.8. If 24 students are
randomly selected, find the probability that the mean of their test score is less than 76.
Ans. 0.97
Chapter 7
14. Find the margin of error and construct the CI corresponds to: 90% confidence level, n = 130,
x = 65. Ans. 0.428 < p < 0.572
15. Use the given data to find the minimum sample size required to estimate the population
proportion: E = 0.03, confidence level = 94%, p̂ = 0.13.
Ans. 469
Construct a CI for the population mean . Assume that the population has a normal distribution.
16. A saving and loan association needs information concerning the checking account balances
of its local customers. A random sample of 14 accounts was checked and yielded a mean
balance of $664.14 and a standard deviation of $297.29. Find a 98% confidence interval for
the true mean checking account balance for local customers. Ans. $453.59 <  < $874.69
Chapter 8 Hypothesis Testing
Provide an appropriate response.
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17. In a population, 11% of people are left handed. In a simple random sample of 160 people
selected from this population, the proportion of left handers is 0.10. What is the number of
left handers in the sample and what notation is given to that number? What are the values of
p and p̂ ? Ans. 16 left handers in the sample, x = 16, p = 0.11, p̂ = 0.10
18. In a sample of 47 adults selected randomly from one town, it is found that 9 of them have
been exposed to a particular strain of the flu. Find the p-value for the test of the claim that the
proportion of all adults in the town that have been exposed to this strain of the flu is 8%. Use
the significance level of 0.05.
Ans.claim: p = 0.08
Alternative: p  0.08
Hypothesis test: H0: p = 0.08
H1: p  0.03
9
 0.08
pˆ  p
Test statistic: z =
 2.8174
 47
pq
0.08(0.92)
n
47
P-value: = 2  Normalcdf(2.8174, ∞, 0,1)  0.0048 < 0.05
Reject H0
Final conclusion: There is sufficient evidence to warrant rejection of the claim that p = 0.08.
19. In tests of computer component, it is found that the mean time between failures is 520 hours.
A modification is made which is supposed to increase the time between failures. Tess on a
random sample of 10 modified components resulted in the following times (in hours)
between failures: (mean is 537.1, sd = 20.7)
518
548
561
523
536
499
538
557
528
563
At the 0.05 significance level, test the claim that for the modified components, the mean time
between failures is greater than 520 hours. Use the P-value method.
Ans.
claim:  > 520
Alternative:  ≤ 520
Hypothesis test: H0:  = 520
H1:  > 520
x   537.1  520
Test statistic: t =
 2.612

s
20.7
n
10
P-value: 0.014
Reject H0
Final conclusion: there is sufficient evidence to support the claim that  > 520.
Chapter 9
20. Given sample sizes and number of successes to find the pooled estimate p . Round your
answer to the nearest thousandth.
n1 = 34, x1 = 15; n2 = 414, x2 = 105. Ans. (x1 + x2)/(n1 + n2) = 120/448  0.268
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21. Assume that you plan to use a significance level of  = 0.05 to test the claim that p1 = p2. Use
the given sample sizes and numbers of successes to find the P-value for the hypothesis test.
n1 = 100, x1 = 41; n2 = 140, x2 = 35.
Ans. p1 = 0.41; p2 = 35/140 = 0.25; p = 76/240  0.317;
claim: p1 = p2
Alternative: p1  p2
Hypothesis test: H0: p1 = p2
H1: p1  p2 (Two tail test)
Test statistic z  2.626;
P-value = 2Normalcdf(2.626, 10000, 0, 1)  0.0086 < 0.05
Reject H0
Final conclusion: The is sufficient evidence to warrant rejection of the claim that
p1 = p2.
22. Test the indicated claim about the means of two populations. Assume that all requirements
are met. Do not assume that 1 = 2. A researcher was interested in comparing the response
times of two different cab companies. Company A and B were each called at 50 randomly
selected times. The calls to company A were made independently of calls to company B. The
response time for each call were recorded. The summary statistics were as follows:
Company A
Company B
Mean response time
7.6 minutes
6.9 minutes
Standard deviation
1.4 minutes
1.7 minutes
Use a 0.02 significance level to test the claim that the mean response time for company A is
the same as the mean response time for company B. Use the P-value method.
Ans. claim: 1 = 2
Alternative: 1  2
Hypothesis test: H0: 1 = 2
H1: 1  2 (two tail test)
 = 0.02
Critical value = 2.403
x x
7.6  6.9
Test statistic t = 1 2 
 2.248
s12 s22
1.4 2 1.7 2


n1 n2
50
50
P-value = 0.029 (> 0.02)
Fail to reject H0
Final conclusion: there is not sufficient evidence to support the claim that 1 = 2.
(The following will be worth 1/4 of the final)
23. The two data sets are dependent. Find d to the nearest tenth.
A
70
67
56
63
51
B
22
24
29
25
22
Ans. 37.0
5
24. Assume that all requirements are met. The table below shows the weights of seven subjects
before and after following a particular diet for two months.
subject
A
B
C
D
E
F
G
before
180
188
172
193
195
168
158
after
173
179
179
198
181
170
146
Using a 0.01 level of significance, test the claim that the diet is effective in reducing weight.
Ans. Claim: d > 0
Alternative: d ≤ 0
Hypothesis test: H0: d = 0
H1: d > 0 (original claim) (right tail)
 = 0.01
d 0 40
Test statistic: t =
 1.327

sd
8.52
n
8
Critical Value = 2.998
Fail to reject H0.
There is not sufficient evidence to support the claim that the diet is effective in reducing
weight.
Chapter 10
25. Given that LCC r = –0.844, and n = 5. Find the critical values of r and determine whether or
not the given r represents a significant linear correlation. Using  = 0.05.
Ans. critical values = 0.878. Since | r | < 0.878, the given r does not represent a significant
correlation.
26. Describe the error in the stated conclusion. Given: there is no significant linear correlation
between scores of a math test and scores on a verbal test. Conclusion: there is no relationship
between scores on the math test and scores on the verbal test.
Ans. there can be other relations (e.g. non-linear)
27. Use the given data to find the best predicted value of the response variable. Eight pairs of
data yield r = 0.708 and the regression equation ŷ = 55.8 + 2.79x. Also, y = 71.125. What is
the best predicted value of y for x = 5.7?
Ans. since r > 0.707 (at  = 0.05). The data are linearly correlated. So the best predicted
value of y is 55.8 + 2.79(5.7)  71.7.
28. Use the given data to find the equation of the regression line. Round the final values to three
significant digits, if necessary.
hours 0
3
4
5
12
Score 8
2
6
9
12
Ans. ŷ = 4.88 + 0.525x
29. Using the same data as in the above problem. Find the residual at the sample point (3, 2)
Ans. Residual  2 – (4.88 + 0.5253)  –4.455 (residual = observed – predicted)
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